In a country popular for train travel, you have planned some train travelling one year in advance. The days of the year that you will travel is given as an array days. Each day is an integer from 1 to 365.

Train tickets are sold in 3 different ways:

a 1-day pass is sold for costs[0] dollars;
a 7-day pass is sold for costs[1] dollars;
a 30-day pass is sold for costs[2] dollars. The passes allow that many days of consecutive travel. For example, if we get a 7-day pass on day 2, then we can travel for 7 days: day 2, 3, 4, 5, 6, 7, and 8.

Return the minimum number of dollars you need to travel every day in the given list of days.

Example 1:

Input: days = [1,4,6,7,8,20], costs = [2,7,15]
Output: 11
Explanation: 
For example, here is one way to buy passes that lets you travel your travel plan:
On day 1, you bought a 1-day pass for costs[0] = $2, which covered day 1.
On day 3, you bought a 7-day pass for costs[1] = $7, which covered days 3, 4, ..., 9.
On day 20, you bought a 1-day pass for costs[0] = $2, which covered day 20.
In total you spent $11 and covered all the days of your travel.

Example 2:

Input: days = [1,2,3,4,5,6,7,8,9,10,30,31], costs = [2,7,15]
Output: 17
Explanation: 
For example, here is one way to buy passes that lets you travel your travel plan:
On day 1, you bought a 30-day pass for costs[2] = $15 which covered days 1, 2, ..., 30.
On day 31, you bought a 1-day pass for costs[0] = $2 which covered day 31.
In total you spent $17 and covered all the days of your travel.

Note:

1 <= days.length <= 365
1 <= days[i] <= 365
days is in strictly increasing order.
costs.length == 3
1 <= costs[i] <= 1000

Solution

Time complexity : O(n)
Space complexity : O(n)

class Solution {
public:
    int mincostTickets(vector<int>& days, vector<int>& costs) {
        int last_day = days.back();
        vector<bool> activate(last_day+1);
        vector<int> dp(last_day+1);
        for (int day: days) activate[day]=true;
        dp[0] = 0;
        
        for (int i=1; i<=last_day; ++i) {
            if (!activate[i]) {
                dp[i] = dp[i-1];
                continue;
            }
            dp[i] = dp[i-1]+costs[0];
            dp[i] = min(dp[i], dp[max(0, i-7)]+costs[1]);
            dp[i] = min(dp[i], dp[max(0, i-30)]+costs[2]);
        }
        
        return dp.back();
    }
};

dp[i]代表到第i天為止，可以花的最少的錢。
用activate記錄哪些天要用到票。
若第i天要用到票，則
dp[i] = min(dp[i-1]+cost[0], dp[i-7]+cost[1], dp[i-30]+cost[2])
買一天的票，買七天的票，買三十天的票，三者取最小。
最後回傳dp.back()，到最後一天為止，可以花的最少的錢。